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Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 38 maths textbook solution.

Answers (1)

Answer    :  \frac{1}{2}

Hint         : use indefinite integral formula and the limits to solve this integral

Given      \int_{0}^1{}\frac{1-x^2}{(1+x^2)^2}dx

Solution :\int_{0}^1{}\frac{1-x^2}{(1+x^2)^2}dx

=-\int_{0}^{1} \frac{x^{2}\left(1-\frac{1}{x^{2}}\right)}{x^{2}\left(\frac{1}{x}+x\right)^{2}} d x=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}} d x
put x+1x=t⇒1-1x2dx=dt
when x=0 then t= ∞ ,when x=1 then t=2

Therefore ,

\begin{aligned} &\int_{0}^{1} \frac{1-x^{2}}{\left(1+x^{2}\right)^{2}} d x \\ &=-\int_{\infty}^{2} \frac{1}{t^{2}} d t=\int_{2}^{\infty} t^{-2} d t \mid \\ &=\left[\frac{t^{-2+1}}{-2+1}\right]_{2}^{\infty}=\left[\frac{t^{-1}}{-1}\right]_{2}^{\infty} \end{aligned}



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